Asked by kk2
A tire 0.650 m in radius rotates at a constant rate of 240 rev/min. Find the speed of a small stone lodged in the tread of the tire (on its outer edge). (Hint: In one revolution, the stone travels a distance equal to the circumference of its path, 2¥ðr.)
What is the acceleration of the stone?
and its direction?
a)away from the center of the path
b)toward the center of the path
c)tangent to the path
What is the acceleration of the stone?
and its direction?
a)away from the center of the path
b)toward the center of the path
c)tangent to the path
Answers
Answered by
drwls
Angular speed of wheel = 240*(2 pi)/60 = 25.13 rad/s = w
Speed of stone (relative to axle) = R*w
The speed of the stone relative to the ground depends upon its height above the ground.
Centripetal acceleration of stone = R*w^2 = 411 m/s^2, , toward the center.
Speed of stone (relative to axle) = R*w
The speed of the stone relative to the ground depends upon its height above the ground.
Centripetal acceleration of stone = R*w^2 = 411 m/s^2, , toward the center.
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